Difference between revisions of "2024 AMC 10B Problems/Problem 8"

Latest revision as of 13:55, 14 November 2024

Problem

Let $N$ be the product of all the positive integer divisors of $42$ . What is the units digit of $N$ ?

$\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8$

Solution 1

The factors of $42$ are: $1, 2, 3, 6, 7, 14, 21, 42$ . Multiply unit digits to get $\boxed{\textbf{(D) } 6}$

Solution 2

The product of the factors of a number $n$ is $n^\frac{\tau(n)}{2}$ , where $\tau(n)$ is the number of positive divisors of $n$ . We see that $42 = 2^1 \cdot 3^1 \cdot 7^1$ which has $(1+1)(1+1)(1+1) = 8$ factors, so the product of the divisors of $42$ is

$42^\frac{8}{2} = 42^4.$

But we only need the last digit of this, which is the same as the last digit of $2^4$ . The answer is $\boxed{\textbf{(D) } 6}$ .

Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)

https://youtu.be/QLziG_2e7CY?feature=shared

~ Pi Academy

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=24EZaeAThuE

2024 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 20: / Line 20: @@
 ~ Pi Academy
+==Video Solution 2 by SpreadTheMathLove==
+https://www.youtube.com/watch?v=24EZaeAThuE
 ==See also==
 {{AMC10 box|year=2024|ab=B|num-b=7|num-a=9}}
 {{MAA Notice}}

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